a)
for the puck :
F = force applied in the direction of pull
N = normal force on the puck in upward direction by the surface of table
W = weight of the puck in down direction due to force of gravity
b)
along the vertical direction , normal force balance the weight of the puck , hence the net force is same as the force of pull F .
so F = ma where m = mass of puck , a = acceleration
Fnet = F
c)
since the net force acts in the direction of force of pull F , hence the puck accelerates in the same direction .
To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,
I = q*N
Here q is Flow of electric charge in one second and N the number of electron flow per second.
A the same time the power is described as the applied voltage for the current.
P = VI
We know the charge of electron,
Coulombs, then the current is


And the power in the Beam is



Given data
ball throws upwards at an angle 60°
Horizontal component (Vh) = 12.5 m/s,
Vertical component (Vv) = 21.7 m/s ,
The magnitude of throw/resultant velocity (V) = ?
The resultant velocity /the velocity with which ball is throws is determined by the following equation
V = √[(Vh)² + (Vv)²]
= √[(12.5)² + (21.7)²]
= 25.04 m/s
<em> The resultant velocity or the velocity with which the ball is thrown is 25 m/s</em>
The electrons are removed from the rod and transferred to the silk cloth. This leaves the rode positively charged and the cloth negatively charged.